Question: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $40.8$ years; the standard deviation is $6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living longer than $28.8$ years.
Answer: $40.8$ $34.8$ $46.8$ $28.8$ $52.8$ $22.8$ $58.8$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $40.8$ years. We know the standard deviation is $6$ years, so one standard deviation below the mean is $34.8$ years and one standard deviation above the mean is $46.8$ years. Two standard deviations below the mean is $28.8$ years and two standard deviations above the mean is $52.8$ years. Three standard deviations below the mean is $22.8$ years and three standard deviations above the mean is $58.8$ years. We are interested in the probability of a bear living longer than $28.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the bears will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the bears will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $28.8$ years and the other half $({2.5\%})$ will live longer than $52.8$ years. The probability of a particular bear living longer than $28.8$ years is ${95\%} + {2.5\%}$, or $97.5\%$.